A square and a rectangle have equal areas. If their perimeters are $p_1$ and $p_2$ respectively, then

### Concept & Optimization Principle This is a standard geometric property: for any given, fixed area, the quadrilateral with the minimum perimeter is a square. As a shape gets more elongated (like a long rectangle), its perimeter increases for the same area. ### Step-by-Step Solution * **Step 1:** Let the area of both the square and the rectangle be $A$. * **Step 2:** For the square, side $a = \sqrt{A}$. Its perimeter $p_1 = 4\sqrt{A}$. * **Step 3:** For the rectangle, let dimensions be $l$ and $b$ such that $l \times b = A$. Its perimeter is $p_2 = 2(l + b)$. * **Step 4:** By the AM-GM inequality, for positive numbers $l$ and $b$: $\frac{l + b}{2} \ge \sqrt{lb}$. * **Step 5:** Substituting $A$: $\frac{l + b}{2} \ge \sqrt{A} \implies l + b \ge 2\sqrt{A} \implies 2(l + b) \ge 4\sqrt{A}$. * **Step 6:** So, $p_2 \ge p_1$. Since a rectangle (not a square) has $l \neq b$, the inequality is strict: $p_2 > p_1$, or $p_1 < p_2$. ### Exam Strategy & Shortcut Use a counter-example. Let Area $= 16$. Square: side $= 4$, Perimeter $p_1 = 4 \times 4 = 16$. Rectangle: sides $= 8$ and $2$. Perimeter $p_2 = 2(8 + 2) = 20$. Clearly, $16 < 20$, so $p_1 < p_2$. ### Common Pitfall Students often confuse this with the opposite rule: "For a fixed perimeter, the square maximizes area." Memorizing these rules backwards leads to picking the wrong inequality sign. ### Final Answer Therefore, the correct answer is **$p_1 < p_2$**.